Slide25.JPG

Slide26.JPG
Slide27.JPG Slide28.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 33 Find intervals in which the function given by f(𝑥) =3/10 𝑥4 – 4/5 𝑥^3– 3𝑥2 + 36/5 𝑥 + 11 is (a) strictly increasing (b) strictly decreasingf(𝑥) = 3/10 𝑥4 – 4/5 𝑥^3– 3𝑥2 + 36/5 𝑥 + 11 Finding f’(𝒙) f’(𝑥) = 3/10 × 4𝑥^3 – 4/5 × 3𝑥^2 – 3 × 2x + 36/5 + 0 f’(𝑥) = 12/10 𝑥^3– 12/5 𝑥^2– 6x + 36/5 f’(𝑥) = 6/5 𝑥^3− 12/5 𝑥^2– 6x + 36/5 f’(𝑥) = 6(𝑥^3/5−(2𝑥^2)/5−𝑥+6/5) f’(𝑥) = 6((𝑥^3 − 2𝑥^2− 5𝑥 + 6)/5) = 6/5 (𝑥^3−2𝑥^2−5𝑥+6) = 6/5 (𝑥−1)(𝑥2−𝑥−6) = 6/5 (𝑥−1)(𝑥2−3𝑥+2𝑥−6) = 6/5 (𝑥−1)[𝑥(𝑥−3)+2(𝑥−3)] = 6/5 (𝑥−1)(𝑥+2)(𝑥−3) Hence, f’(𝒙) = 𝟔/𝟓 (𝒙−𝟏)(𝒙+𝟐)(𝒙−𝟑) Putting f’(𝒙) = 0 𝟔/𝟓 (𝒙−𝟏)(𝒙+𝟐)(𝒙−𝟑) = 0 f’(𝑥) = 6((𝑥^3 − 2𝑥^2− 5𝑥 + 6)/5) = 6/5 (𝑥^3−2𝑥^2−5𝑥+6) = 6/5 (𝑥−1)(𝑥2−𝑥−6) = 6/5 (𝑥−1)(𝑥2−3𝑥+2𝑥−6) = 6/5 (𝑥−1)[𝑥(𝑥−3)+2(𝑥−3)] = 6/5 (𝑥−1)(𝑥+2)(𝑥−3) Hence, f’(𝒙) = 𝟔/𝟓 (𝒙−𝟏)(𝒙+𝟐)(𝒙−𝟑) Putting f’(𝒙) = 0 𝟔/𝟓 (𝒙−𝟏)(𝒙+𝟐)(𝒙−𝟑) = 0 (𝑥−1)(𝑥+2)(𝑥−3) = 0 Hence, x = –2 , 1 & 3 Plotting points on number line Hence, f(𝑥) is strictly decreasing on the interval 𝑥 ∈ (−∞,−𝟐)& (𝟏 , 𝟑) f(𝑥) is strictly increasing on the interval 𝑥 ∈ (−𝟐,𝟏) & (𝟑 , ∞)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.